Solve this

$\sin \theta+\cos \theta=\frac{1}{2}$

$\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=\frac{1}{4}$

$\sin 2 \theta=-\frac{3}{4}$

Now :

$\cos 4 \theta=1-2 \sin ^{2} 2 \theta$

$=1-2\left(-\frac{3}{4}\right)^{2}$

$=1-2 \times \frac{9}{16}=-\frac{1}{8}$

$\sin 6 \theta=3 \sin 2 \theta-4 \sin ^{3} 2 \theta$

$=\left(3-4 \sin ^{2} 2 \theta\right) \cdot \sin 2 \theta$

$=\left[3-4\left(\frac{9}{16}\right)\right] \cdot\left(-\frac{3}{4}\right)$

$\Rightarrow\left[\frac{3}{4}\right] \times\left(-\frac{3}{4}\right)=-\frac{9}{16}$

$16[\sin 2 \theta+\cos 4 \theta+\sin 6 \theta]$

$16\left(-\frac{3}{4}-\frac{1}{8}-\frac{9}{16}\right)=-23$